c. tsang fermat numbers(2010)

7/30/2019 C. TSANG Fermat Numbers(2010)

http://slidepdf.com/reader/full/c-tsang-fermat-numbers2010 1/24

Fermat Numbers

Cindy Tsang

University of Washington

Math414 Number Theorey

Professor William SteinMarch 12, 2010



1

0 Content

1 Introduction2

2 Background of Fermat Numbers 3

3 Geometric Interpretation of Fermat Numbers 5

4 Factoring Status of Fermat Numbers 6

5 Basic Properties of Fermat Numbers 7

6 Primality of Fermat Numbers 12

7 Mersenne Numbers and Fermat Numbers 17

8 Infinitude of Fermat Primes 19

9 Divisibility of Fermat Numbers 21

10 References 23



2

1 Introduction

Prime numbers are widely studied in the field of number theory. One approach to investigate

prime numbers is to study numbers of a certain form. For example, it has been proven that there

are infinitely many primes in the form a + nd , where d ≥ 2 and gcd(d , a) = 1 (Dirichlet’s

theorem). On the other hand, it is still an open question to whether there are infinitely many

primes of the form n2

+ 1.

In this paper, we will discuss in particular numbers of the form 2+ 1 where n is a

nonnegative integer. They are called Fermat numbers, named after the French mathematician

Pierre de Fermat (1601 – 1665) who first studied numbers in this form. It is still an open problem

to whether there are infinitely many primes in the form of 2+ 1. We will not be able to answer

this question in this paper, but we will prove some basic properties of Fermat numbers and

discuss their primality and divisibility. We will also briefly mention numbers of the form 2n – 1

where n is a positive integer. They are called Mersenne numbers, named after the French

mathematician Marin Mersenne. In section6, we will see how Mersenne numbers relate to the

primality of Fermat numbers.

Pierre de Fermat (1601 – 1665) Marin Mersenne (1588 – 1648)[pictures from http://en.wikipedia.org/Pierre_de_Fermat & http://en.wikipedia.org/wiki/Marin_Mersenne]



3

2 Background of Fermat Numbers1

Fermat first conjectured that all the numbers in the form of

2

+ 1 are primes. However, in

1732, Leonhard Euler refuted this claim by showing that F 5 = 232

+ 1 = 4,294,967,297

= 641 x 6,700,417 is a composite. It then became a question to whether there are infinitely many

primes in the form of 2+ 1. Primes in this form are called Fermat primes. Up-to-date there are

only five known Fermat primes. (See section4 for more details on the current status of Fermat

numbers.)

Leonhard Paul Euler (1707 – 1783) Carl Friedrich Gauss (1977 – 1855)[pictures from http://en.wikipedia.org/wiki/Euler & http://en.wikipedia.org/wiki/Gauss]

In 1796, the German mathematician Carl Friedrich Gauss (1977 – 1855) found an interesting

relationship between the Euclidean construction (i.e. by ruler and compass) of regular polygons

and Fermat primes. His theorem is known as Gauss’s Theorem.

Gauss’s Theorem2 . There exists an Euclidean construction of the regular n-gon if and only if

n = 2i p1 p2··· p j, where n ≥ 3, i ≥ 0, j ≥ 0, and p1 , p2,…, p j are distinct Fermat primes.

1 All historical information in this section is from Reference1 Chapter1.2 A proof of Gauss’s Theorem can be found in Reference1 Chapter16.



4

Gauss’s theorem implies that all 2n-gons for n ≥ 2 are constructible. Moreover, since so far

only five Fermat numbers are known to be prime, it implies that for n odd, there are only

5C 1 + 5 C 1 + 5C 1 + 5C 1 + 5C 1 = 31 n-gons that are known to be Euclidean constructible. If it turnsout that there is only a finite number of Fermat primes, then this theorem would imply that there

is only a finite number of Euclidean constructible n-gons for n odd. The figure below shows five

Euclidean constructible n-gons.

Triangle, pentagon, heptadecagon, 257-gon and 65537-gon.[figure from Reference1 Chapter4]



5

3 Geometric Interpretation of Fermat Numbers

As Gauss’s theorem suggests, Fermat numbers might be closely related to some of the

problems in Geometry. It is hence useful if we can understand what they mean geometrically.

A Fermat number F n = 2+ 1 (for n ≥ 1) can be thought of as a square whose side length is

2plus a unit square (see figure1). Hence, determining whether a (Fermat) number is a

composite or not is equivalent to determining whether we can rearrange the unit-square blocks to

form a rectangle (see figure2). Moreover, determining whether an integer d divides a (Fermat)

number is the same as deciding whether we can reorganize the blocks to form a rectangle with

base d ; or alternatively, we can also think of it as determining whether we can “fill” the area with

a number of r·d unit-square blocks for some integer r (see figure3).

Figure1. F 2 = 42

+ 1 = 17

Figure2. F 2 = 17 is not a composite because no matter

how you rearrange the blocks, you cannot get a rectangle.

Figure3. F 2 = 17 is not divisible by 3.

Some of the properties we will prove in section5 can be easily understood if we interpret

them geometrically. The reader should pay close attention. We will also make remarks on several

of them.



6

4 Factoring Status of Fermat Numbers3 (as of February 3, 2010)

The below table only shows the factoring status of Fermat numbers up to n = 200. For larger

Fermat numbers and other details, see http://www.prothsearch.net/fermat.html#Summary.

Prime

Composite with no known factors

Composite with complete factorization

Composite with incomplete factorization

Unknown

0

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70

71 72 73 74 75 76 77 78 79 80

81 82 83 84 85 86 87 88 89 90

91 92 93 94 95 96 97 98 99 100

101 102 103 104 105 106 107 108 109 110111 112 113 114 115 116 117 118 119 120

121 122 123 124 125 126 127 128 129 130

131 132 133 134 135 136 137 138 139 140

141 142 143 144 145 146 147 148 149 150

151 152 153 154 155 156 157 158 159 160

161 162 163 164 165 166 167 168 169 170

171 172 173 174 175 176 177 178 179 180

181 182 183 184 185 186 187 188 189 190

191 192 193 194 195 196 197 198 199 200

Total number of F n primes known 5

Largest F n prime known F 4 = 65537

Total number of F n composites known 243

Largest ten F n composites known F 476624, F 495728, F 567233, F 585042, F 617813,

F 67205, F 960897 , F 2145351, F 2167797 , F 2478792

3 All information from this section is from Reference2 http://www.prothsearch.net/fermat.html#Summary



7

5 Basic Properties of Fermat Numbers

In this section, we will prove some basic properties of Fermat numbers.

Theorem14 . For n ≥ 1, F n = ( F n-1 – 1)

2+ 1.

Proof. ( F n-1 – 1)2

+ 1 = (2+ 1 – 1)

2+ 1 = 2

+ 1 = F n □

Remark1. This theorem is obvious if we interpret it geometrically:

Figure4. Any Fermat number F n is exactly a square with side length F n-1 – 1 plus a unit square.

Theorem25 . For n ≥ 1, F n = F 0··· F n-1 + 2.

Proof. We will prove this by induction.

When n = 1, we have F 0 + 2 = 3 + 2 = 5 = F 1.

Now assume F n = F 0··· F n-1 + 2.

Then, F 0··· F n + 2 = F 0··· F n-1· F n + 2

= ( F n – 2)· F n + 2 (induction hypothesis)

= (2 – 1)·( 2

+ 1) + 2

= 2+ 1 = F n+1 □

4 Theorem is found in Reference3. Proof is due to the author.5 Theorem is found in Reference3. Proof is due to the author.



8

Remark2. To understand the proof of Theorem2 geometrically, we can think of F n – 2 as a

square with side length F n-1 – 1 minus a unit square (see figure5). It is divisible by

F n-1 = 2

+ 1 because we can form a rectangle by moving the top row and make it a columnon the right (see figure6). To see that it is also divisible by F n-k for 2 ≤ k ≤ n, we can use the

induction hypothesis that F n-k divides F n-1 – 2 = 2– 1. It means that we can fill each column

of the rectangle in figure5 evenly by r·F n-k number of blocks for some integer r (see figure7).

Figure5. A 2x 2

square Figure6. A (2– 1) x (2

+ 1) rectangle

minus a unit square

Figure7. Each column can be filled evenly by F n-k .

Corollary2.1. [Reference1, p.27] For n ≥ 1, F n ≡ 2 (mod F k ) for all k = 0, 1, … , n – 1.

Proof. It is equivalent to say that F k | F n – 2, which is implied by Theorem2. □

Corollary2.2. [Reference1, p.28] For n ≥ 2, the last digit of F n is 7.



9

Proof. It follows directly from Corollary2.1 that F n ≡ 2 (mod 5). Since all Fermat numbers are

odd, it follows that F n ≡ 7 (mod 10). □

Corollary2.3. No Fermat number is a perfect square.

Proof. F 0 = 3 and F 1 = 5 are obviously not a perfect square. For F n where n ≥ 2, by Corollary2.2,

F n ≡ 7 (mod 10). But only numbers that are congruent to 0, 1, 4, 5, 6, or 9 (mod 10) can be a

perfect square. □

Remark2.3. This is quite intuitive if we think of F n as a square plus a unit square block. You

can’t possibly rearrange the block to form a perfect square.

Corollary2.4. [Reference1, p.31] Every Fermat number F n for n ≥ 1 is of the form 6m – 1.

Proof. It is equivalent to show that F n + 1 is divisible by 6. From Theorem2, we have

F n + 1 = 3·F 1··· F n + 2 + 1 = 3·( F 1··· F n + 1), where F 1··· F n + 1 is an even number. □

Theorem36 . For n ≥ 2, F n = F

2n-1 – 2·( F n-2 – 1)

2.

Proof. F 2n-1 – 2·( F n-2 – 1)

2= (2

+ 1)2 – 2·(2

– 1 + 1)2

= 2+ 2·2+ 1 – 2·2

= 2+ 1 = F n □

6 Theorem is found in Reference3. Proof is due to the author.



10

Theorem47 . For n ≥ 2, F n = F n-1 +2

· F 0···F n-2.

Proof. We will prove this by induction.

When n = 2, we have F 1 + 2

2

· F 0 = 5 + 2

2

·3 = 17 = F 2.

Now assume F n = F n-1 +2· F 0···F n-2.

Then, F n + 2· F 0···F n-1 = F n + 2

·(2· F 0···F n-2)· F n-1

= F n + 2· F n-1·( F n – F n-1) (induction hypothesis)

= 2+ 1 + 2

·(2+ 1)·(2

– 2)

= 2+ 1 + 2

·(2+ 1)· 2

·( 2– 1)

= 2+ 1 + 2

·(2– 1)

= 2+ 1 + 2 – 2

= 2+ 1 = F n+1 □

Theorem5. [Reference1, p.28] For n ≥ 2, every Fermat number has infinitely many

representations in the form x2

– 2y2, where x and y are both positive integers.

Proof. First, from Theorem3, ( x0 , y0) = ( F n-1 , F n-2 – 1) gives one such representation. Now

notice that (3 x + 4 y)2

– 2·(2 x + 3 y)2

= 9 x2

+ 24 xy + 16 y2

– 8 x2

– 24 xy – 18 y2

= x2

– 2 y2. If x and y

are both positive, then 3 x + 4 y > x and 2 x + 3 y > y are also positive. This means that we can find

( xi , yi) recursively by setting ( xi , yi) = (3 xi-1 + 4 yi-1, 2 xi-1 + 3 y-1). The set of all points ( xm , ym) we

find will be infinite, and each point will give a representation for F n in the desired form. □

Theorem6. [Reference3] No two Fermat numbers share a common factor greater that 1.

Proof. Assume for contradiction that there exist F i and F j such that a > 1 divides both of them.

Also, without loss of generality, assume that F j > F i.

From Theorem4, we know that F j = F j-1 + · F 0···F i···F j-2. Since a divides F i and F j, a alsodivides F j-1 and hence F 0···F i···F j-1. Then, a has to divide the difference F j – F 0··· F j-1, which

equals 2 by Theorem2. It follows that a = 2, but all Fermat numbers are obviously odd. □

7 Theorem is found in Reference3. Proof is due to the author.



11

We can in fact use Theorem6 to prove that there are infinitely many primes. Although there

are already proofs about the infinitude of primes without using the concept of Fermat numbers, it

is interesting and worthwhile to see an alternative proof.

Corollary6. [Reference3] There are infinitely many primes.

Proof. Define a sequence { pi} in the following way:

i) if F i is a prime, then define pi = F i;

ii) if F i is a composite, then define pi = a prime factor of F i.

All the pi’ s are distinct by Theorem6. Hence, the set { pi : i = 1, 2, 3 …} contains infinitely many

primes. □

Theorem7. [Reference1, p.29] No Fermat number F n for n ≥ 2 can be expressed as the sum of

two primes.

Proof. Assume for contradiction that there exists n ≥ 2 such that F n could be expressed as the

sum of two primes. Since F n is odd, one of the primes must be 2. Then the other prime would

equal F n – 2 = 2 – 1 = (2

+ 1)·(2 – 1), which is not a prime. □



12

6 Primality of Fermat Numbers

Recall that we have defined Fermat numbers to be numbers in the form of

2

+ 1 where n is

a nonnegative integer. There is actually another definition for Fermat numbers, namely numbers

in the form of 2n

+ 1 where n is a nonnegative integer. We have chosen the former definition

because it seems to be more commonly used and it gives the properties that we have proved

earlier. Notice that Theorem6 is false if we had chosen the other definition. A counterexample is

21

+ 1 = 3 and 23

+ 1 = 9 have a common factor 3.

However, if we are only interested in Fermat numbers that are primes, then it does not matter

which definition we use, as we will see from the next theorem.

Theorem8. [Reference3] If 2n

+ 1 is a prime, then n is a power of 2.

Proof. Suppose n is a positive integer that is not a power of 2. Then we can write n = 2r · s for

some nonnegative integer r and some positive odd integer s. Also recall the identity

an

– bn

= (a – b)·(an-1

+ an-2

b + ··· + abn-1

+ bn-1

),

which implies that a – b divides an

– bn. Now substituting a = 2

r , b = – 1 and n = s, we have 2

r + 1

divides 2rs

– (–1) s

= 2n

+ 1. However, r < n, which means that 2n

+ 1 is not a prime. Hence, n

must be a power of 2 in order for 2n

+ 1 to be a prime. □

The next two theorems concern the properties of Fermat primes.

Theorem9. [Reference1, p. 31] No Fermat prime can be expressed as the difference of two pth

powers, where p is an odd prime.

Proof. Assume for contradiction that there is such a Fermat prime. Then, F n = a p

– b p

= (a – b)·(a p-1 + a p-2b + ··· + ab p-1 + b p-1), where a > b and p is an odd prime. Since F n is a prime, it

must be the case that a – b = 1. Moreover, by Fermat’s Little Theorem, a p ≡ a (mod p) and b p ≡b

(mod p). Thus, F n = a p

– b p

≡ a – b = 1 (mod p). This implies p | F n – 1 = 2, which is

impossible because the only integer that divides 2is 2. □



13

Theorem10. [Reference1, p.30] The set of all quadratic nonresidues of a Fermat prime is equal to

the set of all its primitive roots.

Proof. First let a be a quadratic nonresidue of the Fermat prime F n and let e = ord F na. According

to Fermat’s little theorem, a( F n – 1)

≡ 1 (mod F n), so e | F n – 1 = 2. It follows that e = 2

k for

some nonnegative integer k ≤ 2n. On the other hand, by Euler’s criterion, / =

≡ – 1(mod F n). Hence, if k < 2n, then 2

k | 2 and so ≡ 1 (mod F n), which is a

contradiction. So, k = 2n

and ord F na = 2. Therefore, a is a primitive root modulo F n.

Conversely, suppose r is a primitive root modulo F n. It follows that r ( F n – 1)/2

≠ 1 (mod F n), and

so by Euler’s criterion, r cannot be a quadratic residue. □

Now recall that Fermat’s little theorem can be used to test whether a number is a prime or not.

However, it does not work for Fermat numbers if we choose the base to be 2, as we will see in

Theorem11.

Lemma11. [Reference1, p.36] For m ≤ 2n

– 1, F m |

2

– 2.

Proof. 2 – 2 = 2(2 – 1) = 2(2+ 1 – 2) = 2( – 2) = 2 F 0··· (Theorem2). □

Theorem11. [Reference1, p.36] All Fermat numbers are primes or pseudoprimes to base 2.

Moreover, if 2n

+ 1 is a pseudoprime to the base 2, then n is a power of 2.

Proof . Since n ≤ 2n

– 1, from Lemma11, F n | 2 – 2. Since F n ≠ 2, we have F n | 2 – 1, which

is equivalent to say that 2 ≡ 1 (mod F n).

Now suppose 2n

+ 1 is a pseudoprime to the base. Then we have

2≡ 1 (mod 2

n+ 1). Notice

that 2n ≡ –1 (mod 2

n+ 1), so 2

2n ≡ 1 (mod 2

n+ 1). Now let e = (2). First e ≥ n + 1 for

otherwise we will have 2e ≤ 2

n< 2

n+ 1. Moreover, e | 2n, so it follows that e = 2n. But e | 2

,

which is only possible if n is a power of 2. □



14

Theorem11 is not true, however, for other bases in general. For examples, F 5 = 4294967297

is not a pseudoprime to base 5 or 6, since we have 54294967296

≡ 2179108346 (mod 4294967297)

and 6

4294967296

≡

3029026160 (mod 4294967297). Hence, it is still possible to use Fermat’s littletheorem to test the primality of a Fermat number as long as we do not choose 2 to be our base.

Other than Fermat’s little theorem, there are various other primality tests that can be used to

test whether a Fermat number (or any number in general) is a prime or not. In the rest of this

section, we will discuss in particular two of them. The first one is called Selfridge’s test

(Theorem13), and the second one is a generalized version of Pepin’s test (Theorem14). The

proof for the latter involves the use of the Jacobi symbol, which we will introduce here.

Definition. [Reference1, p.25] Let a be an integer and suppose n ≥ 3 is an odd integer. Write

n = p1· p2··· pr , where the pi’s are odd primes but not necessarily distinct. Then the Jacobi symbol

is defined by

= ∏

, where

is the Legendre symbol.

The Jacobi symbol has properties very similar to those of the Legendre symbol. We state six

of them in the following theorem. To familiarize ourselves with the Jacobi symbol, we will prove

two of the properties stated. The reader should verify the rest himself. They can be proved using

the properties of the Legendre symbol.



15

Theorem12. [Reference1, p.25] Let m > 1 and n > 1 be odd integers and let a and b be integers.

Then we have the following:

i) if a ≡ b (mod n), then =

;

ii) =

;

iii) =

;

iv) = 1,

= 1/;

v) = 1/;

vi) = 1/

.

Proof 8 . Unless specified, all the in this proof represent the Legendre symbol.

i) If a ≡ b (mod n), then a ≡ b (mod pi) for all prime factors pi of n. Hence,

=

for

all pi’ s and ∏

= ∏

. Thus, the Jacobi symbols and

; are equal. □

ii) The Jacobi symbol = ∏

= ∏

= ∏

∏

=

, where

and

are the Jacobi symbols. □

Theorem13. [Reference1, p.42] Let N > 1 and let the prime-power factorization of N – 1 be

∏ . Then N is a prime if and only if for each prime pi where i = 1, 2, …, r , there exists an

integer ai > 1 such that ai N-1

≡ 1 (mod N ) and ai(N-1)/pi

≠ 1 (mod N ).

Proof. If N is a prime, then there exists a primitive root a that satisfies both conditions.

Conversely, it suffices to show that Φ( N ) = N – 1. Let ei = ord N ai. Then ei | N – 1 but

ei ( N – 1)/ pi. Hence, pik i | ei. We also have ai

Φ( N ) ≡ 1 (mod N ) by Euler’s theorem, so ei | Φ( N ).

Consequently, pik i | Φ( N ) for all i = 1, 2, …, r , and hence, N – 1 | Φ( N ). But Φ( N ) ≤ N – 1, so

Φ( N ) = N – 1. □

8 Proof is due to the author.



16

Theorem14. [Reference1, p.42] For n ≥ 2, the Fermat number F n is prime if and only if

a( F n – 1)/2

≡ –1 (mod

F n), where a is an integer such that the Jacobi symbol

= –1 for all n ≥ 2.

Proof. First assume that F n is prime. Then the Jacobi symbol is just the Legendre symbol.

So by Euler’s criterion, a( F n – 1)/2

≡

≡ –1 (mod

F n).

Now assume that the congruence holds. Then we have both a( F n – 1)/2

≡ –1 (mod

F n) and

a( F n – 1)

≡ 1 (mod

F n). Since 2 is the only prime factor of F n – 1, by Theorem13, F n is a prime. □

Corollary14. [Reference1, p.42] For n ≥ 2, the Fermat number F n is prime if and only if

3( F n – 1)/2

≡ –1 (mod

F n).

Proof. It suffices to show that

= –1. From Corollary2.1, F n ≡ 2 (mod 3). Moreover, since

F n ≡ 1 (mod 4), by Theorem12,

= (–1)

(3 – 1)(4k + 1 – 1)/4 =

= = –1. □



17

7 Mersenne Numbers and Fermat Numbers

Recall that we have defined Mersenne numbers to be numbers of the form 2n – 1 where n is a

positive integer. Some definitions require n to be a prime. However, like the case of Fermat

numbers, if we are only interested in Mersenne numbers that are primes, then it does not matter

which definition we choose. We can see that in the following theorem.

Theorem15. [Reference4] A Mersenne number M n = 2n – 1 is prime only if n is a prime.

Proof. Recall the identity 2ab

– 1 = (2a

– 1)·(1 + 2a

+ 22a

+ ··· + 2(b-1)a

). Hence if n = ab is not a

prime, then M n = 2n – 1 is divisible by 2

a– 1 ≠ 1. □

The next two theorems show how Mersenne numbers relate to the primality of the associated

Fermat numbers.

Lemma16. [Reference1, p.44] If p is a prime, then all Mersenne numbers M p are prime or

pseudoprimes to the base 2.

Proof. Let M p = 2 p

– 1 be a Mersenne number where p is a prime. If M p is a composite, then p is

odd. By Fermat’s little theorem, ( M p – 1)/2 = 2 p-1

– 1 ≡ 0 (mod p). So ( M p – 1)/2 = kp for some

positive integer k . Hence, M p = 2 p

– 1 | 2kp

– 1 = 2( M p – 1)/2

– 1. It is equivalent to say that

2( M p – 1)/2

≡ 1 (mod M p), which implies that 2 M p – 1

≡ 1 (mod M p). □

Theorem16. [Reference1, p.45] Let p be a prime such that p ≡ 3 (mod 4). Then the Fermat

number F p is prime if and only if M p( F p – 1)/2

≡ – 1 (mod F p), where M p is the associated Mersenne

number.

Proof. By Theorem14, it suffices to show that

= – 1.

By Lemma16, 2 ≡ 1 (mod M p), and multiplying 2 to both sides we get 2 ≡ 2

(mod M p). This implies that F p = 2·2 + 1 ≡ 5 (mod M p). Moreover, since p ≡ 3 (mod 4),

M p = 2 p

– 1= 24k +3

– 1 = 8·24k

– 1 ≡ 3·1 – 1 = 2 (mod 5). Thus, by Theorem 12,

=

=

= =

= –1. □



18

Theorem17. [Reference1, p.45] Let p be a prime such that p ≡ 3 or 5 (mod 8). Then the Fermat

number F p is prime if and only if M p( F p+1 – 1)/2

≡ – 1 (mod F p+1), where M p is the associated

Mersenne number.

Proof. Again by Theorem14, it sufficies to show that

= – 1. By the same argument in Theorem16, we can show that F p ≡ 5 (mod M p). Then by Theorem1,

F p+1 = ( F p – 1)2

+ 1 = 42

+ 1 = 17 (mod M p). First we assume p ≡ 3 (mod 8), then M p = 28k +3

– 1 =

8·162k

– 1 ≡ 8 – 1 = 7 (mod 17). Hence,

=

=

= =

= –1.

Now if we assume p ≡ 5 (mod 8), then M p = 28k +5

– 1 ≡ 25

– 1 = –3 ≡ 14 (mod 17). Hence,

=

=

= –1. □



19

8 Infinitude of Fermat Primes

As we have noted before, there are only five known Fermat primes so far. In fact, it has been

shown that F n is composite for 5 ≤ n ≤ 32 and many other larger n (from section4). Whether

there is an infinite number of Fermat primes is still an open question, and below shows a

heuristic argument that suggests there is only a finite number of them. This argument is to due to

Hardy and Wright [Reference1, p.158].

There is only a finite number of Fermat primes.

Recall that the Prime Number Theorem says π( x) ~

, where π( x) is the number of primes ≤ x.

Hence π( x) < for some constant A, and the probability that x is a prime is at most

.

For x = 2+ 1, the probability that it is a prime is ≤

.

Hence, the expected number of primes in this form is ∑ 2A which is a finite number.

We can use the same reasoning to argue that there are infinitely many twin primes.

There are infinitely many twin primes.

Recall the Prime Number Theorem can be stated using limit: lim

/.

Hence give ε > 0, there exists a number X such that 1 – ε <

/ for all x > X.

Thus, the probability that n and n+2 are both primes is

·

·

1 1 for n > X.

So the expected number of twin primes is > ∑ 1 ∑

1 which diverges.

However, we must be careful that there two arguments do not prove that there are really only

finitely many Fermat primes or infinitely many twin primes. After all, they are only heuristic, as

we can see in a similar argument below.



20

There are infinitely many primes in the form of 2n

+ 1.

Using the exact same argument as above, the expected number of primes in this form is

∑

1 ∑

1 which diverges.

But we know from Theorem8 that the sets {2n

+ 1: it is a prime} and {2+ 1: it is a prime}

are the same set. This latter argument suggests Hardy and Wright’s argument does not take into

account of the properties of Fermat numbers. It is to say that the variable x is not that random. It

works largely because gaps between successive Fermat numbers are extremely large.

Nevertheless, given any number (even a number of a particular form), it is more likely to be a

composite than prime. Therefore, bounding the probability of it being a prime by a lower boundgives a weaker argument that bounding it from above.



21

9 Divisibility of Fermat Numbers

In the last two sections, we focused on the primality of Fermat numbers and the properties of

Fermat primes. However, if a Fermat number is found to be composite, we are interested in what

its factorization is, or at least, what properties do its divisors have to have. We will end our

discussion of Fermat numbers in this section by proving several theorems about their divisors.

Theorem 18. [Reference1, p.37] Let q = pm be a power of an odd prime p, where m ≥ 1. Then the

Fermat number F n is divisible by q if and only if ordq2 = 2n+1

.

Proof. First suppose q | F n, then q | (2+ 1)·(2

– 1) = 2– 1, and hence 2

≡ 1 (mod q).

It follows that 2n+1 = k ordq2 for some positive integer k . Thus, k is a power of 2 and so is ordq2.

Let e = ordq2 = 2 j. If j < n + 1, then we have q | 2

– 1 = 2– 1. But this is impossible

because q | 2+ 1 and q ≠ 2. Hence, j = n + 1 and so ordq2 = 2

n+1.

Conversely, if we assume that ordq2 = 2n+1

, then q | 2– 1 = (2

+ 1)·( 2– 1). Since q

is an odd prime, q divides either 2+ 1 or 2

– 1. But q cannot divide 2– 1 because

2n < ordq2. Hence q | 2

+ 1 = F n. □

Theorem19(Euler). [Reference1, p. 38] If p is a prime and p | F n, then p is of the form

p = k2n+1

+ 1 , where k is a positive integer.

Proof. By Fermat’s little theorem, 2 p-1

≡ 1 (mod p), and it follows that ordq2 | p – 1. Hence,

k ordq2 = p – 1 for some positive integer k , and by Theorem18, p = k ordq2 + 1 = k2n+1

+ 1. □



22

Theorem20(Lucas). [Reference1, p.59] If n > 1 and a prime p divides F n, then p is of the form

p = k 2n+2

+ 1, where k is a positive integer.

Proof. Let b = 2

(2

– 1). Since p | F n = 2

+ 1, we have 2 ≡ –1 (mod p). Hence,

b2

= 2(2

– 2·2+ 1) ≡ 2

(–1 – 2·2+ 1) = –2·2

≡ 2 (mod p). It then follows

that = 2

≡ –1 (mod p) and thus ≡ 1 (mod p). Consequently, e = ord pb = 2

jfor some

j ≤ n + 2. If j < n + 2, then – 1 =

– 1 ≡ 2– 1 ≡ 0 (mod p). This contradicts to the

previous result that 2+ 1 ≡ 0 (mod p). Hence, j = n + 1 and ord pb = 2

n+2.

Now since b2≡ 2 (mod p), it follows that gcd(b, p) =1. By Fermat’s little theorem, b

p-1 ≡ 1

(mod p). Thus, ord pb = 2n+2

| p – 1, and hence p = k 2n+2

+ 1 for some positive integer k . □

Corollary20. [Reference1, p.39] If n > 1, then any divisor d > 1 of a Fermat number F n is of the

form k 2n+2

+ 1, where k is a positive integer.

Proof. Consider the product (k 2n+2

+ 1)·(k 2m+2

+ 1). Without loss of generality, assume m ≥ n.

Then (k 2n+2

+ 1)·(k 2m+2

+ 1) = k 22

m+n+4+ k 2

n+2+ k 2

m+2+ 1 = (k

22

m+2+ k + k 2

m-n)2

n+2, which is

also in the form of k 2n+2

+ 1. Following from Theorem19, all divisors have the form k 2n+2

+ 1. □



23

10 References

1. M. Krizek, F. Luca and L. Somer, 17 Lectures on Fermat Numbers – From Number Theory

to Geometry, Springer-Verlag, New York, 2001.

2. W. Keller, Prime factors k·2n

+ 1 of Fermat numbers F m and complete factoring status.

http://www.prothsearch.net/fermat.html#Summary

3. _____, Fermat number

http://en.wikipedia.org/wiki/Fermat_numbers

4. _____, Mersenne number

http://en.wikipedia.org/wiki/Mersenne_numbers

c. tsang fermat numbers(2010)

Documents